Question

1. how many grams of solute are in 250. ml of a 0.10 m cacl₂ solution?

ans 1) 0.040 m 2) 0.25 m 3) 0.20 m 4) 0.75 m 5) 44 g 6) 2.8 g

Explanation:

Step1: Recall the formula for molarity

Molarity ($M$) is defined as moles of solute ($n$) per liter of solution ($V$ in liters), so $M = \frac{n}{V}$. We can rearrange this to find moles: $n = M \times V$.
First, convert the volume from milliliters to liters. The volume is 250. mL, and since 1 L = 1000 mL, $V = \frac{250.}{1000} = 0.250$ L.

Step2: Calculate moles of $\ce{CaCl2}$

Given $M = 0.10$ M and $V = 0.250$ L, use $n = M \times V$. So $n = 0.10 \, \text{mol/L} \times 0.250 \, \text{L} = 0.025$ mol.

Step3: Find the molar mass of $\ce{CaCl2}$

The molar mass of Ca is 40.08 g/mol, and for Cl it's 35.45 g/mol. For $\ce{CaCl2}$, the molar mass ($MM$) is $40.08 + 2 \times 35.45 = 40.08 + 70.90 = 110.98$ g/mol (approx 111 g/mol).

Step4: Calculate mass of solute

Mass ($m$) is moles times molar mass, so $m = n \times MM$. Substituting $n = 0.025$ mol and $MM = 110.98$ g/mol, we get $m = 0.025 \, \text{mol} \times 110.98 \, \text{g/mol} \approx 2.77$ g, which rounds to 2.8 g.

Answer:

1. 2.8 g