Question

Amount in g of simple containing 80% NaOH required to prepare 60 litre of 0.5 M solution is :

• A. 1000
• B. 1200
• C. 1500
• D. 1600

Answer 1

Answer: (C)

1500

Volume of solution $=60$ litres

Molarity of solution $=0.5$M

Using the formula,

Molarity $=$ No. of moles of solutes / Volume of solutions in litre.

$0.5=$ No. of moles / $60$.

No. of moles of $NaOH=30$moles.

molar mass of $NaOH=40$g/mole.

no. of mole $=\frac{mass}{molarmass}$

mass $=$ (no. of moles) (molar mass)

mass $=$ $\left(30\right)\left(40\right)$

$\therefore$ mass of $NaOH=1200$grams

Now, let the mass of solution be $xg.$

According to question,

mass of $NaOH=80$% of solution.

$1200=\left(80/100\right)\times \left(x\right)$

$x=1500$g

$\therefore$ mass of sample or solution is $1500$ grams.

Therefore, the option is C.