Chemistry MCQs for NEET — Practice Questions with Answers

$\left(\text{Weight percentage}=\frac{\text{Weight of solute}}{\text{Weight of solution}}\times 100\right)$

Total weight of solution = (15 + 35) g = 50 g

Weight percentage of methyl alcohol $=\frac{\text{Weight\hspace{0.17em}of\hspace{0.17em}methyl\hspace{0.17em}alcohol}}{\text{Weight\hspace{0.17em}of\hspace{0.17em}solution}}\times 100=\frac{15}{50}\times 100=30\%$ 

Part per million $=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times {10}^6=\frac{5.8\times {10}^{-3}g}{{10}^{3}g}\times {10}^6$ = 5.8 ppm  

ppm of F^– ions $=\frac{\text{Mass of solute}}{\text{Mass of solution}}\times {10}^6=\frac{0.2}{500}\times {10}^6$ = 400 ppm 

Eq. wt. of $H_{2}S{O}_4=\frac{\text{Mol. mass of }{H}_{2}S{O}_4}{\text{Basicity of }{H}_{2}S{O}_4}=\frac{98}{2}=49$

∴ Number of g equivalent of H₂SO₄ $=\frac{\text{Weight in }g}{\text{Eq. mass}}=\frac{9.8}{49}=0.2$

250 cm³ of solution contain H₂SO₄= 0.2 g equivalent

∴ 1000 cm³ of the solution contain H₂SO₄ $=\frac{0.2}{250}\times 1000g$ equivalent = 0.8 g equivalent

Hence normality of the solution = 0.8 N  

Wt. of solute $=\frac{N\times V\times geq.wt.}{1000}=\frac{0.5\times 200\times 40}{1000}=4g$ 

$N_1{V}_1+N_2{V}_2+N_3{V}_3=N_4{V}_4$

$50\times 10+25\times 12+40\times 5=N_4\times 1000$ or N₄ = 1 N 

100 ml of 0.3 N HCl contains $HCl=0.03g$eq.,

200 ml of 0.6 N H₂SO₄ contains H₂SO₄ $=\frac{0.6}{1000}\times 200=0.12g$eq.

Total g eq. = 0.15, Total volume = 300 ml Finally normality $=\frac{0.15}{300}\times 1000=0.5$

Alternatively $N_1{V}_1+N_2{V}_2=N_3{V}_3$

i.e., $0.3\times 100+0.6\times 200=N_3\times 300$

or $0.3+1.2=3N_3$

or $N_3=1.5/3=0.5$ 

Normality of oxalic acid solution $=\frac{6.3}{63}\times \frac{1000}{250}=0.4$

$N_1{V}_1=N_2{V}_2$

$0.1\times V_1=0.4\times 10$ or $V_1=40ml$   

Weight of base (w) = 10.6 g; g eq. wt. of base = 53; Vol. of acid (V) = 100 ml; Normality of acid (N) = ?

$\frac{w}{geq.wt.}=\frac{V\times N}{1000}$; $\frac{10.6}{53}=\frac{100\times N}{1000}$; $N=\frac{1000\times 10.6}{100\times 53}=2$  

Consider 1000 ml of water

Mass of 1000 ml of water $=1000\times 1=1000\text{\hspace{0.17em}}g$

Number of moles of water $=\frac{1000}{18}=55.5$

Molarity $=\frac{\text{No.\hspace{0.17em}of moles of water}}{\text{Volume in litre}}=\frac{55.5}{1}=55.5M$