Question

How many grams of sodium hydroxide must be added to make $200{\text{ mL}}$ of a $1{\text{ M NaOH}}$ solution?
A. $8{\text{ g}}$
B. $16{\text{ g}}$
C. $40{\text{ g}}$
D. $80{\text{ g}}$

Answer 1

You can calculate the number of moles by dividing weight with molecular weight. You can calculate the molarity by dividing the number of moles with volume in liter.

Complete answer:
Let $x{\text{ g}}$ of sodium hydroxide must be added to make $200{\text{ mL}}$ of a $1{\text{ M NaOH}}$ solution.
Convert the unit of volume from milliliter to liter by dividing with 1000.
$\Rightarrow\dfrac{{200{\text{ mL}}}}{{1000{\text{ mL/L}}}} = 0.200{\text{L}}$
The molecular weight of sodium hydroxide is ${\text{40 g/mol}}$
Divide mass of sodium hydroxide with its molecular weight to obtain the number of moles of sodium hydroxide
$\Rightarrow\dfrac{{x{\text{ g}}}}{{{\text{40 g/mol}}}}{\text{ = }}\dfrac{x}{{{\text{40}}}}{\text{moles }}$
Divide the number of moles of sodium hydroxide with its volume to obtain its molarity.
$\Rightarrow\dfrac{{\dfrac{x}{{{\text{40}}}}{\text{moles }}}}{{0.200{\text{ L}}}}{\text{ = }}\dfrac{x}{{0.200 \times {\text{40}}}}{\text{M = }}\dfrac{x}{{\text{8}}}{\text{M }}$
But the molarity of sodium hydroxide solution is ${\text{1M}}$
$\Rightarrow\dfrac{x}{{\text{8}}}{\text{M = 1M}}$
$\Rightarrow x{\text{ = 8 g}}$
Hence, ${\text{8 grams}}$ of sodium hydroxide must be added to make $200{\text{ mL}}$ of a $1{\text{ M NaOH}}$ solution?

The correct option is the option A.

Note: When you multiply the molarity with volume (in Litre), you get the number of moles. When you multiply the number of moles with molecular weight you get the mass. So you can use the alternate approach to solve the same problem as shown below.
$\Rightarrow1{\text{ M}} \times \left( {\dfrac{{200{\text{ mL}}}}{{1000{\text{ mL/L}}}}} \right) = 0.200{\text{ mol}}$
$\Rightarrow0.200{\text{ mol}} \times 40{\text{ g/mol}} = 8{\text{ g}}$