Question

solutions and molarity

1. find the mass to prepare a 3.0% solution of sucrose in 30 ml of di water.

Explanation:

Step1: Recall mass - percent formula

Mass - percent = $\frac{\text{mass of solute}}{\text{mass of solution}}\times100\%$. For a dilute aqueous solution, we assume the density of water is 1 g/mL, so the mass of 30 mL of water is 30 g. Let the mass of sucrose be $m$. The mass of the solution is $(m + 30)$ g.

Step2: Substitute values into formula

We know that the mass - percent is 3.0% or 0.03. So, $0.03=\frac{m}{m + 30}$. Cross - multiply gives $0.03(m + 30)=m$.

Step3: Expand and solve for $m$

Expanding, we get $0.03m+0.9 = m$. Rearranging terms: $m-0.03m=0.9$, $0.97m = 0.9$, and $m=\frac{0.9}{0.97}\approx0.9$ g (if we assume the mass of the solute is much smaller than the mass of the solvent and approximate the mass of the solution as the mass of the solvent, $0.03=\frac{m}{30}$, then $m = 0.03\times30=0.9$ g).

Answer:

0.9 g