Class 11th

1.36. The molar mass of MnO₂is 87g and the molar mass of HCl is 36.5 g.

According to the equation, 4 moles of HCl (i.e. 4 x 36.5g = 146 g) reacts with 1 mol of MnO₂ (87g).

So, for 5.0 g of MnO₂ will react with:

= 5 x146/87 = 8.40 g of MnO₂

Therefore, 8.4 g of HCl will react with 5 g of MnO₂.

1.35. Step 1: 0.75 M HCl means 0.75 mol per 1000 mL or 0.75 x 36.5 g in 1000 mL.

i.e. 1000 mL of 0.75 M HCl contains 0.75 x 36.5 g HCl

Therefore, 25 mL of 0.75 M HCl contains 0.75 x 36.5 x 25/ 1000 g = 0.6844 g

Step 2: To calculate mass of CaCO₃reacting completely with 0.6844 g of HCl
CaCO₃  (s) + 2HC1 (aq)———>CaCl₂ (aq) +CO₂ (g) + H₂O
2 mol of HCl, i.e., 2 x 36.5 g = 73 g HCl react completely with CaC0₃ = 1 mol = 100 g

Therefore, 0.6844 g HCl will react completely with CaCO₃ = 100/73 x 0.6844 g = 0.938 g

1.33.  (i) 52 moles of Ar

Ans:1 mole of Ar = 6.022 * 10²³ atoms of Ar

Therefore, 52 mole of Ar = 52 * 6.022 * 10²³ atoms of Ar= 3.131 * 10²⁵ atoms of Ar

(ii) 52 u of He

Ans:1 atom of He = 4u of the He

Or, 4 u of He = 1 atom of He

So, 52 u of He = 52/4 atom of He = 13 atoms of He.

(iii) 52 g of He

Ans:4g of He = 6.022 * 10²³ atoms of He

So, 52g of He = 6.022 * 1023 * 52/4 atoms of He

= 7.8286 * 10²⁴ atoms of He

1.32. Molar mass of argon is

= [ (35.96755 * 0.337/100)+ (37.96272 * 0.063/100)+ (39.9624 * 99.60/100)]g mol^-l

= [0.121+0.024+39.802] g mol^-l

= 39.947 g mol^-l

So, the molar mass of argon is 39.947 g/ mol.

1.31. First we need to find the least precise number to find the significant figures.

(i) Least precise number of the calculation  is 0.112

Therefore, number of significant figures in the answer = Number of significant figures in the least precise number, i.e. 3

(ii) Least precise number of calculations = 5.364

Therefore, number of significant figures in the answer will be = Number of significant figures in 5.364 = 4

(iii) 0.0125+0.7864+0.0215

Since the least number of decimal places in each term is four, the number of significant figures in the answer will also be 4.

1.30. 1 mol of 12C atoms = 6.02 x 10²³ atoms = 12 g

Or, 6.02 x 10²³ atoms of 12C have mass = 12 g

Therefore, 1 atom of 12C will have mass = 12 x 1/6.02 x 10²³ = 1.9927 x 10^-23 g

1.29. According to the formula of mole fraction,

X (C₂H₅OH) = 0.040 = n (ethanol) / [n (ethanol) + n (water)]

Now, n (water) = 1000g/18g mol^-1 = 55.55 moles                                          [? Density of water=1kg m^-3]

Therefore, 0.040 = n (ethanol) / [n (ethanol) + 55.55]

i.e., 0.04 x n (ethanol) + 2.222 = n (ethanol)

i.e., 2.222 = [1- 0.04] n (ethanol)

i.e., n (ethanol) = 2.222 / 0.96 = 2.314 M

1.28. The number of atoms in each of the given elements are calculated as below:

(i) 1 g Au = 1 / 197 mol = 1/ 197 x 6.02 x 10²³ atoms

(ii) 1 g Na = 1/ 23 mol = 1/ 23 x 6.02 x 10²³ atoms

(iii) 1 g Li = 1/ 7 mol = 1 / 7 x 6.02 x 10²³ atoms

(iv) 1 g of Cl₂ = 1/ 71 mol = 1/ 71 x 6.02 x 10²³ atoms

Therefore, since the denominator is the smallest in case of Li, 1 g Li has the largest number of atoms

1.27.  (i) 28.7 pm = 28.7 x 10^-12 m = 2.87 x 10^-11 m

(ii) 15.15 µs = 15.15 x 10^-6 s = 1.515 x 10^-5 s

(iii) 25365 mg = 25365 mg x 10^-6 kg = 2.5365 x 10^-2 kg