Question

 If 7.5 g of sodium nitrate, NaNO3 is dissolved in 85ml of H2O (density of H2O = 1.0g/ml), calculate the %(wt/wt) concentration of H2O in the solution. 

Given:

 Solution:

 A brand of rubbing alcohol says, it contains 70% (vol/vol) isopropyl alcohol. How many mL of isopropyl alcohol are there in 600 ml of the solution in the bottle? 

Given:

 Solution:

 The label of betadine skin cleanser says 7.5 % solution. Taking it to be % (wt/vol), how many grams of betadine (Providone-Iodine) are present in 50 mL bottle? 

Given:

 Solution:

 Calculate the mole fraction of sulfuric acid (H2SO4) in 8% (wt/wt) aqueous H2SO4 solution (molar masses: H2SO4 = 98 g/mol, H2O = 18 g/mol 

Given:

 Solution:

 Calculate the molar concentration of a solution that contains 23 g of potassium hydroxide. KOH in 250 ml of solution. Molar mass of KOH is 56 g/mol. 

Given: 

Solution:



Answer 1

Part 1

Mass of water $=85×1=85g$

$\%$ concentration of water $=\frac{85}{(85+7.5)}×100\%=91.89\%$

Part 2

Ml of isopropyl alcohol $=\frac{70}{100}×600=420ml$

Part 3

$\frac{Mass}{50}=\frac{7.5}{100}\implies$ mass $=3.75g$

Part 4

Mass of H₂SO₄ $=8g$

Mass of water $=100-8=92g$

Moles of H₂SO₄ $=\frac{8}{98}=0.08163$

Moles of water $=\frac{92}{18}=5.11111$

Mole fraction $=\frac{0.08163}{(5.11111+0.08163)}=0.01572$

Part 5

Mole concentration $=\frac{23}{56}×\frac{1000}{250}$

$=1.643M$