Question

how many grams of barium iodide, bai2, must be dissolved to prepare 250. ml of a 0.151 m aqueous solution of the salt? use the references to access important values if needed for this question. 3 item attempts remaining

Explanation:

Step1: Recall the molar - mass formula

First, find the molar mass of $BaI_2$. The molar mass of $Ba$ is approximately $137.33\ g/mol$, and the molar mass of $I$ is approximately $126.90\ g/mol$. Since there are 2 iodine atoms in $BaI_2$, the molar mass of $BaI_2$, $M = 137.33+2\times126.90=137.33 + 253.8=391.13\ g/mol$.

Step2: Use the molar - concentration formula

The molar - concentration formula is $n = C\times V$, where $n$ is the number of moles, $C$ is the molarity, and $V$ is the volume in liters. Given $C = 0.151\ M$ and $V=250.0\ mL=0.2500\ L$. Then $n = 0.151\ mol/L\times0.2500\ L = 0.03775\ mol$.

Step3: Calculate the mass

The mass $m$ of a substance is given by $m=n\times M$. Substitute $n = 0.03775\ mol$ and $M = 391.13\ g/mol$ into the formula. So $m=0.03775\ mol\times391.13\ g/mol\approx14.7\ g$.

Answer:

$14.7$