Question

a saturated solution of a salt was made by adding 36.00 g to 150.0 g water. there was 5.00 g of salt on the bottom that didnt dissolve. what is the % solubility of the salt ?
the mass of the salt that dissolved is 1 g.
2 g
% solubility = --------- 100% = 3 %
4 g
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. 15 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 dissolved salt

The mass of salt added is \(36.00\) g and the mass of undissolved salt is \(5.00\) g. So, the mass of dissolved salt \(m = 36.00 - 5.00=31.00\) g.

Step2: Identify the formula for percentage solubility

The formula for percentage solubility is \(\%\text{ solubility}=\frac{\text{mass of solute (dissolved)}}{\text{mass of solvent}}\times100\%\). Here, the mass of solute (dissolved) is \(31.00\) g and the mass of solvent (water) is \(150.0\) g.

Step3: Calculate the percentage solubility

\(\%\text{ solubility}=\frac{31.00}{150.0}\times 100\%=\frac{3100}{150}\% = 20.67\%\) (rounded to two decimal places)

Answer:

1. C. \(31.00\)
2. C. \(31.00\)
3. T. \(20.67\)
4. D. \(150.0\)