Question

15.00 g of naoh were added to 185.00 g of water.

1. what is the % m/m?

1 g naoh
% m/m = ---------------- 100% = 2 % m/m
3 g soln

1. how many grams of naoh in 444.44 g of the solution? (dont use the % here,

use actual numbers)
4 g naoh
(5 g soln)(----------------) = 6 g naoh
7 g soln
a. 36.00 b. 5.00 c. 31.00 d. 150.0 e. 12 f. 350.00 g. 100.0
h. 2.4 x 10^-2 i. 1.2 x 10^-3 j. 1.786 k. 7.143 l. 25.00 m. 7.500
n. 125.0 o. 500.0 p. 15.00 q. 185.00 r. 444.44 s. 2.00 x 10^3
t. 20.67 u. 0.500 v. 499.5 w. 30.0 x. 33.33 y. 1.00 x 10^5
z. 10.0 aa. 1.00 bb. 10^6 cc. 6.00 dd. 6.00 x 10^-5
ee. 6.00 x 10^6 ff. 20.0 gg. 200 hh. 1.20 x 10^6 ii. 60
jj. 0.100 kk. 4.00 ll. 10^3 mm. 0.0667 nn. 66.7

Explanation:

Step1: Calculate the mass of the solution

The mass of the solution is the sum of the mass of the solute (NaOH) and the mass of the solvent (water).
$$15.00 + 185.00=200.00$$

Step2: Calculate the mass - mass percentage ($\%m/m$)

The formula for mass - mass percentage is $\%m/m=\frac{\text{mass of solute}}{\text{mass of solution}}\times100\%$.
Substitute the values: $\frac{15.00}{200.00}\times100\% = 7.500\%$

Step3: Calculate the mass of NaOH in 444.44g of the solution

We know that the ratio of NaOH to the solution is $\frac{15.00}{200.00}$.
Let $x$ be the mass of NaOH in 444.44g of the solution. Then $x = 444.44\times\frac{15.00}{200.00}$
$$x=\frac{444.44\times15.00}{200.00}=33.333\approx33.33$$

Answer:

1. \(1 = P\) (15.00g NaOH), \(3 = G\) (200.00g soln), \(2 = M\) (7.500% m/m)
2. \(4 = P\) (15.00g NaOH), \(5 = R\) (444.44g soln), \(7 = G\) (200.00g soln), \(6 = X\) (33.33g NaOH)