a 0.2449 g sample of a monoprotic acid neutralizes 21.41 ml of a 0.09369 m koh solution. calculate the molar…

Explanation:

Step1: Determine moles of KOH

Use the formula $n = M\times V$. Given $M = 0.09369\ M$ and $V=21.41\ mL = 0.02141\ L$. $n_{KOH}=0.09369\ mol/L\times0.02141\ L$ $n_{KOH}= 0.09369\times0.02141 = 0.0020068029\ mol$

Step2: Relate moles of acid and KOH

For a monoprotic acid ($HA$) reacting with $KOH$: $HA + KOH=KA + H_2O$. The mole - ratio of $HA$ to $KOH$ is 1:1. So $n_{acid}=n_{KOH}=0.0020068029\ mol$.

Step3: Calculate molar mass of acid

Use the formula $M=\frac{m}{n}$. Given $m = 0.249\ g$ and $n = 0.0020068029\ mol$. $M=\frac{0.249\ g}{0.0020068029\ mol}\approx124.0\ g/mol$

Answer:

$124.0\ g/mol$