Solutions MCQs for NEET — Chemistry Questions with Answers

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

$\underset{0.1\mathrm{M}}{{\mathrm{AgNO}}_3}+\underset{0.2\mathrm{M}}{\mathrm{NaCl}}\stackrel{}{\to }\mathrm{AgCl}+{\mathrm{NaNO}}_3$

$0.1\mathrm{M}{\mathrm{AgNO}}_3$ reacts with 0.1 M NaCl to produce

0.1 M AgCl and $0.1\mathrm{M}{\mathrm{NaNO}}_3$

$\therefore {\mathrm{NO}}_3^{-}=\frac{0.1\mathrm{M}}{2}=0.05\mathrm{M}$

[∵ when equal volumes are mixed dilution occurs]  

Molarity $=\frac{\text{Wt.\hspace{0.17em}of\hspace{0.17em}solute}}{\text{Mol.\hspace{0.17em}wt.}}\times \frac{1000}{\text{Vol.\hspace{0.17em}of\hspace{0.17em}solution\hspace{0.17em}(in\hspace{0.17em}}ml.)}=\frac{98}{98}\times \frac{1000}{54.34}=18.4M$

$\left(\text{Vol.\hspace{0.17em}of\hspace{0.17em}solution\hspace{0.17em}}=\frac{{\displaystyle \text{mass}}}{{\displaystyle \text{density}}}=\frac{{\displaystyle 100}}{{\displaystyle 1.84}}=54.34ml\right)$   

Molecular mass of oxalic acid = 126

1000 ml of 1 M oxalic acid require oxalic acid = 126 g

$\therefore \text{\hspace{0.17em}\hspace{0.17em}}250ml$ of 1 M oxalic acid will require oxalic acid $=\frac{126}{1000}\times 250=31.5g$

Hence 250 ml of $\frac{M}{2}$ oxalic acid will require oxalic acid $=31.5\times \frac{1}{2}=15.75g$

∴ Mass of oxalic acid required = 15.75 g.  

In dilution, the following equation is applicable :

$M_1{V}_1$ = $M_2{V}_2$

$10MHCl$ = $5MHCl$

$10\times V_1=5\times 2.00$

$V_1=\frac{5\times 2.00}{10}L=1.00L$