Answer:
Molar mass of the unknown acid is 79 grams
Explanation:
We have to first get moles in 15.0 ml of sodium hydroxide solution:
[tex]{ \sf{1 \: l \: of \:NaOH \: contains \: 0.210 \: moles }} \\ { \sf{0.015 \: l \: of \: NaOH \: contain \: (0.015 \times 0.210) \: moles }} \\ { \underline{ = 0.00315 \: moles \: of \: NaOH}}[/tex]
since mole ratio of acid : base is 1 : 1, so;
moles of acid that reacted is 0.00315 moles of the unknown acid.
then we've to get molar mass:
[tex]{ \sf{0.00315 \: moles \: of \: acid \: weigh \: 0.250 \: g}} \\ { \sf{1 \: mole \: of \: acid \: weighs \: ( \frac{1 \times 0.250}{0.00315} ) \: g}} \\ { \underline{ = 79.4 \: g \approx79 \: grams}}[/tex]