Question

What mass of lead nitrate i.e. $Pb{\left({NO}_3\right)}_2$ (formula weight $=331$), is needed to make $100$ milliliters of a $1.00$ molar solution?

• A. $438$ grams
• B. $33.1$ grams
• C. $269$ grams
• D. $53.8$ grams

Answer 1

Answer: (B)

$33.1$ grams

let $x$ g mass of lead nitrate is required

mole of lead nitrate = $\frac{x}{331}$

then according to question

$molarity$ = $\frac{\text{mole of solute}}{\text{volume of solution}}$

$1M$ = $\frac{x}{331}\times 10$

$x=33.1grams$

Hence correct ans is E