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Chemistry Chemistry Solutions

40 g of glucose (Molar mass = 180) is mixed with 200 mL of water. The freezing point of solution is______ K. (Nearest integer)

[Given : K_f = 1.86 K kg mol^-1; Density of water = 1.00g cm^-3; Freezing point of water = 273.15K].

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alok kumar singh

Contributor-Level 10

Molality (m) = $\frac{(40 / 180)}{0.2} = (\frac{10}{9}) m o l a l$

$Δ T_{f} = T_{f} - T_{f}^{'} = 1.86 * \frac{10}{9} K$

$T_{f}^{'} = 273.15 - 1.86 * \frac{10}{9} K$

$= 271.08 K \approx 271 K$

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New answer posted

a year ago

Chemistry Class 11th

Two flasks I and II shown below are connected by a valve of negligible volume.

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alok kumar singh

Contributor-Level 10

Applying : ${(n_{1} + n_{2})}_{i n i t i a l} = {(n_{1} + n_{2})}_{f i n a l}$

Assuming the system attains a final temperature of T (Such that 300 < T < 60)

(Heat lost by N₂ of container l) = (Heat gained by N₂ of container II)

$n_{1} C m (300 - T) = n_{2} C m (T - 60)$

$(\frac{2.8}{28}) (300 - T) = \frac{0.2}{28} (T - 60)$

14 (300 – T) = T – 60

$P = (\frac{3}{28}) * 8.31 * \frac{284}{3 * 1 0^{- 3}} * 1 0^{- 5} b a r = 0.84287 b a r$

$P = 84.28 * 1 0^{- 2} b a r$

$\approx 84 * 1 0^{- 2} b a r$

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New answer posted

a year ago

Chemistry Class 12th

The number of photons emitted by an monochromatic (single frequency)infrared range finder of power 1 mW and wavelength of 1000 nm, in 0.1 second is x * 10¹³. The value of x is_______. (Nearest integer)

(h = 6.63 * 10^-34 Js, c = 3.00 * 10⁸ ms^-1)

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alok kumar singh

Contributor-Level 10

Energy of emitted photon in 0.1 sec = 10^-4 J

$n * \frac{h c}{λ} = 1 0^{- 4}$

$\frac{n * 6.63 * 1 0^{- 34} * 3 * 1 0^{8}}{1000 * 1 0^{- 9}} = 1 0^{- 4}$

$n = 5.02 * 1 0^{14} = 50.2 * 1 0^{13} \approx 50 * 1 0^{13}$

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New answer posted

a year ago

Chemistry Chemistry Electrochemistry

The resistance of a conductivity cell with cell constant 1.14 cm^-1, containing 0.001 M KCl at 298K is 1500 $Ω .$ The molar conductivity of 0.001 M KCl solution at 298 K in S cm² mol^-1 is__________. (Integer answer)

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alok kumar singh

Contributor-Level 10

$κ = \frac{1}{R} * \frac{l}{A} = [(\frac{1}{1500}) * 1.14] S c m^{- 1} = \frac{1.14}{1500} S c m^{- 1}$

$λ_{m} = \frac{κ}{M} * 1000 S c m^{2} m o l^{- 1}$

$λ_{m} = 1000 * \frac{(\frac{1.14}{1500})}{0.001} S c m^{2} m o l^{- 1} = 760 S c m^{2} m o l^{- 1}$

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New answer posted

a year ago

Chemistry Class 12th

Two flasks I and II shown below are connected by a valve of negligible volume.

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

Applying : ${(n_{1} + n_{2})}_{i n i t i a l} = {(n_{1} + n_{2})}_{f i n a l}$

Assuming the system attains a final temperature of T (Such that 300 < T < 60)

(Heat lost by N₂ of container l) = (Heat gained by N₂ of container II)

$n_{1} C m (300 - T) = n_{2} C m (T - 60)$

$(\frac{2.8}{28}) (300 - T) = \frac{0.2}{28} (T - 60)$

14 (300 – T) = T – 60

$P = (\frac{3}{28}) * 8.31 * \frac{284}{3 * 1 0^{- 3}} * 1 0^{- 5} b a r = 0.84287 b a r$

$P = 84.28 * 1 0^{- 2} b a r$

$\approx 84 * 1 0^{- 2} b a r$

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New answer posted

a year ago

Chemistry Chemistry Electrochemistry

(h = 6.63 * 10^-34 Js, c = 3.00 * 10⁸ ms^-1)

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

Energy of emitted photon in 0.1 sec = 10^-4 J

$n * \frac{h c}{λ} = 1 0^{- 4}$

$\frac{n * 6.63 * 1 0^{- 34} * 3 * 1 0^{8}}{1000 * 1 0^{- 9}} = 1 0^{- 4}$

$n = 5.02 * 1 0^{14} = 50.2 * 1 0^{13} \approx 50 * 1 0^{13}$

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New answer posted

a year ago

Chemistry Chemistry Chemical Bonding and Molecular Structure

The resistance of a conductivity cell with cell constant 1.14 cm^-1, containing 0.001 M KCl at 298K is 1500 $Ω .$ The molar conductivity of 0.001 M KCl solution at 298 K in S cm² mol^-1 is__________. (Integer answer)

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

$κ = \frac{1}{R} * \frac{l}{A} = [(\frac{1}{1500}) * 1.14] S c m^{- 1} = \frac{1.14}{1500} S c m^{- 1}$

$λ_{m} = \frac{κ}{M} * 1000 S c m^{2} m o l^{- 1}$

$λ_{m} = 1000 * \frac{(\frac{1.14}{1500})}{0.001} S c m^{2} m o l^{- 1} = 760 S c m^{2} m o l^{- 1}$

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New answer posted

a year ago

Chemistry Class 11th

Data given for the following reaction is as follows:

$F e O_{(s)} + C_{(g r a p h i t e)} \overset{}{\to} F e_{(s)} + C O_{(g)}$

Substance $Δ_{f} H^{0} (k J m o l^{- 1})$ $Δ S^{0} (J m o l^{- 1} K^{- 1})$

FeO_(s)

...more

Data given for the following reaction is as follows:

$F e O_{(s)} + C_{(g r a p h i t e)} \overset{}{\to} F e_{(s)} + C O_{(g)}$

Substance $Δ_{f} H^{0} (k J m o l^{- 1})$ $Δ S^{0} (J m o l^{- 1} K^{- 1})$

FeO_(s)-266.3 57.49

C_(graphite) 0 5.74

Fe_(s) 0 27.28

CO_(g) -110.5 197.6

The minimum temperature in K at which the reaction becomes spontaneous is______

(Integer answer)

less

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

Minimum temperature at which reaction becomes spontaneous is,

$T_{m i n} = \frac{Δ H^{0}}{Δ S^{0}}$

$Δ {H^{o}}_{m i n} = [Δ H_{f}^{o} (F e) + Δ H_{f}^{o} (C O)] - [Δ H_{f}^{o} (F e O) + Δ H_{f}^{o} (C)]$

$= [0 - 110.5] - [- 266.3 + 0]$

$= 155.8 K J / m o l$

$Δ S ° = [Δ S ° (F e) + Δ S^{o} (C O)] - [Δ S ° (F e O) + Δ S ° (C)]$

$= (27.28 + 197.6) - (57.49 + 5.74) J / m o l K$

$= 161.65 J / m o l K$

$T_{m i n} = \frac{155.8 * 1 0^{3}}{161.65} K = 963.8 K \approx 964 K$

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New answer posted

a year ago

Chemistry Chemistry Some Basic Concepts of Chemistry

100 g of propane is completely reacted with 1000 g of oxygen. The mole fraction of carbon dioxide in the resulting mixture is x * 10^-2. The value of x is _________. (Nearest integer)

[Atomic weight: H = 1.008; C = 12.00; O = 16.00]

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alok kumar singh

Contributor-Level 10

C₃H₈ (g) + 5O₂ (g) -> 3CO₂ (g) + 4H₂O $(l)$

t = 0; 2.27 mol 31.25 mol 0 0

t = $?$ 0 &n

...more

C₃H₈ (g) + 5O₂ (g) -> 3CO₂ (g) + 4H₂O $(l)$

t = 0; 2.27 mol 31.25 mol 0 0

t = $?$ 0 19.9 mol 6.81 mol 9.08 mol

$X_{C O_{2}} = \frac{6.81}{19.9 + 6.81 + 9.08} = 0.1902 = 19.02 * 1 0^{? 2}$

x = 19 (the nearest integer)

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New answer posted

a year ago

Chemistry Class 11th

When 5.1g of solid NH₄HS is introduced into a two litre evacuated flask at 27°C, 20% of the solid decomposes into gaseous ammonia and hydrogen sulphide. The K_p for the reaction at 27°C is x * 10^-2. The value of x is_______. (Integer answer)

[Given R = 0.082 L atm K^-1 mol^-1]

0 Follower 14 Views

alok kumar singh

Contributor-Level 10

Moles of NH₄HS initially taken = $\frac{5.1}{51} = 0.1 m o l$

$N H_{4} H S (S) + N H_{3} (g) + H_{2} S (g)$

t = $\infty$ 0 0.1 mol 0 0

t = $\infty$ 0.1 (1 - 0.2) 0.1 * 0.2 0.1 * 0.2

$P_{N H_{3}} = \frac{n R T}{V} = \frac{0.1 * 0.2 * 0.082 * 300}{2} = 0.246 a t m = P_{H_{2} S}$

$K_{P} = P_{N H_{3}} * P_{H_{2} S} = {(0.246)}^{2} = 0.060516 = 6.05 * 1 0^{- 2}$

x = 6 (Nearest integer).

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