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Q.
The density (in $g \,mL ^{-1}$ ) of a $3.60\, M$ sulphuric acid solution that is $29 \% H _{2} SO _{4}$ (molar mass $=98 \,g \,mol ^{-1}$ ) by mass, will be
Solutions
Solution:
$3.6 \,M$ solution means $3.6$ moles of $H _{2} SO _{4}$ is present in $1000\, mL$ of solution.
$\Rightarrow $ Mass of $3.6$ moles of $H _{2} SO _{4}=3.6 \times 98 \,g =352.8\, g$
$\Rightarrow $ Mass of $H _{2} SO _{4}$ in $1000 \,mL$ of solution $=352.8\, g$
Given, $29 \,g$ of $H _{2} SO _{4}$ is present in $100\, g$ of solution $\left(29\, \% \,H _{2} SO _{4}\right.$ by mass)
$\Rightarrow 352.8\, g$ of $H _{2} SO _{4}$ is present in $\frac{100}{29} \times 352.8=1216 \,g$ of solution
Now, density $=\frac{\text { Mass }}{\text { Volume }}$
$=\frac{1216}{1000}=1.216 \,g / mL =1.22 \,g / mL$