Chemistry Solutions

At $^{}$ $35^{?} C$ , the vapour pressure of ${C S}_{2}$ $_{}$ is $512 m m$ of $H g$ and that of acetone is $344 m m$ of $H g$ . A solution of ${C S}_{2}$ $_{}$ in acetone has a total vapour pressure of $600 m m$ of $H g$ . The false statement among the following is

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Contributor-Level 10

The above mixture will show positive deviation from Raoult's law.

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a month ago

CO? gas is bubbled through water during a soft drink manufacturing process at 298 K. If CO? exerts a partial pressure of 0.835 bar then x m mol of CO? would dissolve in 0.9 L of water. The value of x is ______ (Nearest integer). (Henry's law constant for CO? at 298 K is 1.67*10³ bar)

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Raj Pandey

Contributor-Level 9

P (CO? ) = K? X (CO? )
X (CO? ) = P (CO? )/K? = 0.835 / (1.67 * 10³) = 0.5 * 10? ³
X (CO? ) = n (CO? )/ (n (CO? ) + n (H? O) ≈ n (CO? )/n (H? O) (since n (CO? ) << n (H? O)
n (H? O) in 0.9L = 900g/18gmol? ¹ = 50 mol
n (CO? ) = X (CO? ) * n (H? O) = 0.5 * 10? ³ * 50 = 25 * 10? ³ moles = 25 mmol

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a month ago

1.46 g of a biopolymer dissolved in a 100 mL water at 300 K exerted an osmotic pressure of 2.42 * 10⁻³ bar. The molar mass of the biopolymer is _______ * 10⁴ g mol⁻¹. (Round off to the Nearest integer) [Use: R = 0.083 L bar mol⁻¹ K⁻¹]

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Contributor-Level 10

? = CRT
2.42 * 10? ³ = ( (1.46/M_polymer) / 0.1 ) * 0.083 * 300
M_polymer = (1.46 * 0.083 * 300) / (2.42 * 10? ³ * 0.1)
= 14.96 * 10? gm = 15 * 10? gm

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a month ago

When 3.00 g of substance 'X" is dissolved in 100 g of CCl₄, it raises the boiling point by 0.60K. The molar mass of the substance 'X" is __________g mol⁻¹. (nearest integer)
[Given Kb for CCl₄ is 5.0 K kg mol⁻¹]

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alok kumar singh

Contributor-Level 10

ΔTb = 0.6 K
ΔTb = Kb * m = Kb * (wt. * 1000)/ (mol wt. * Wsolvent (gm)
0.6 = 5 * (3 * 1000)/ (mol wt. * 100)
∴ mol wt. = (15 * 10)/ (0.6) = 1500/6
= 250 g/mole

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2 months ago

224mL of SO_2(g) at 298K and 1 atm passes through 100mL of 0.1M NaOH solution. The non-volatile solute produced is dissolved is 36g of water. The lowering of vapour pressure of solution (assuming the solution is dilute) $(P_{(H_{2} O)}^{o} = 24 m m o f H g) i s x * 1 0^{- 2} m m$ of Hg, the value of x is-----------. (Integer answer)

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Payal Gupta

Contributor-Level 10

Moles of SO₂= $\frac{224 * 1 0^{- 3}}{22.4}$

=0.01 mole

Moles of NaOH = 0.1 * 0.1

= 0.01 mole

SO₂+NaOHNaHSO₃

0.01 mole0.01 mole-

-0.01 mole

Non-volatile solute is NaHSO3

Moles of water = $\frac{36}{18} = 2$

Using ; relative lowering in V.P

$\frac{P^{o} - P}{P^{o}} = i x_{B}$

Where; $Δ P = P^{0} - P$ is lowering in V.P

$Δ P = P^{0} i x_{B}$

i for NaHSO₃ = 2

here; $x_{B} = \frac{n_{B}}{n_{A}}$ since solution is dilute

$Δ P = 24 * 2 * \frac{0.01}{2} = 0.24$

$Δ P = 24 * 1 0^{- 2} m m H g$

So; x = 24

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2 months ago

When 9.45g of ClCH₂COOH is added to 500ml. of water, its freezing point drops by 0.5°C. The dissociation constant of ClCH₂ COOH is x * 10^-3. The value of x is__________ (Rounded off to the nearest integer)

$[k_{f (H_{2} O)} = 1.86 K k g m o l^{- 1}]$

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alok kumar singh

Contributor-Level 10

$Δ T_{f} = 0.5 ° C$

K_f = 1.86

Using, density of water = 1g / mL

i = ?

$Δ T_{f} = i k_{f} . m$

$0.5 = i * 1.86 * \frac{9.45}{94.5 * 500} * 1000$

i = 1.344

Now, using $α = \frac{i - 1}{n - 1}$

n for ClCH₂COOH = 2

a = $\frac{1.344 - 1}{2 - 1} = 0.344$

Using

$K_{a} = \frac{C α^{2}}{1 - α}$

$K_{a} = \frac{0.2 * {(0.344)}^{3}}{0.66} = 36 * 1 0^{- 3}$

So; x = 36

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2 months ago

When 12.2g of benzoic acid is dissolved in 100g of water, the freezing point of solution was found to be -0.93°C $[K_{f} (H_{2} O = 1.86 K k g m o l^{- 1}] .$ The number (n) of benzoic acid molecules associated (assuming 100% association) is…………,

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Contributor-Level 10

$Δ T_{f} = K_{f} * m * i$

0.93 = 1.86 * 1 * i

i = $\frac{1}{2}$ $i = 1 + (\frac{1}{n} - 1)$ $\frac{}{}$

$∴ n = 2$

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2 months ago

C₆H₆ freezes at 5.5°C. The temperature at which a solution of 10g of C₄H₁₀ in 200g of C₆H₆ freeze is………………. °C (The molal freezing point depression constant of C₆H₆ is 5.12°C/m)

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Contributor-Level 10

$Δ T_{f} = k_{f} . m$

$T_{f}^{0} - T_{f} = 5.12 * \frac{(\frac{10}{58})}{(\frac{200}{1000})}$

5.5 – T_f = $\frac{5.12 * 5 * 10}{58}$

$T_{f} = 1.086 ° C = (1.086) ° C ≅ 1 ° C$

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2 months ago

If a compound AB dissociates to the extent of 75% in an aqueous solution, the molality of the solution which shows a 2.5K rise in the boiling point of the solution is……. molal.

(Rounded-off to the nearest integer).

[K_b = 0.52 K kg mol^-1]

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Payal Gupta

Contributor-Level 10

$A B ? A^{+} + B^{-}$

1-x x x $∴ i = 1 + x$

$\begin{array}{l} Δ T_{b} = i * k_{b} * m \\ 2.5 = (1 + x) * 0.52 m \end{array}$

$∴ m o l a l i t y = \frac{2.5}{1.75 * 0.52} = 2.74$

the nearest integer is 3.

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2 months ago