Question

Match the items given in column I with those in column II.

Answer 1

9.8% $H_{2}S{O}_4$ by weight $(density=1.8gm{L}^{-1})$ is 3.6 N and 1.10 m.
$M=\frac{9.8\times 10\times 1.8}{98}=1.8$
$N=1.8\times 2=3.6$
$d=M(\frac{M{w}_2}{1000}+\frac{1}{m})$,
$1.8=1.8(\frac{98}{1000}+\frac{1}{m})\Rightarrow m=1.10$
9.8% $H_{3}P{O}_4$ by weight $(density=1.2gm{L}^{-1})$ is 3.6 N, 1.2 M and 1.10 m.
$M=\frac{9.8\times 10\times 1.2}{98}=1.2$
$N=1.2\times 3=3.6$
$d=M(\frac{M{w}_2}{1000}+\frac{1}{m})$,
$1.8=1.8(\frac{98}{1000}+\frac{1}{m})\Rightarrow m=1.10$
1.8 $N_A$ molecules HCl is 500 mL is 3.6 N. It corresponds to 1.8 equivalents.
$1.8N_{A}molecules=1.8mol$ of HCl 500 mL $=1.8Eq.$
$M=\frac{1.8\times 1000}{500}=3.6M=3.6N$ (n factor $=1)$
250 mL of 4N NaOH + 250 mL of 1.6 M $Ca(OH{)}_2$ corresponds to 1.8 equivalents.
mEq of base $=250\times 4+250\times 1.6\times 2$
$=1000+800$
$=1800mEq$
$=1.8Eq$