Question

What is the volume of water in 150ml of the 35% of sucrose with a specific gravity of 1.115?

Answer 1

Answer:

The volume of the water is 108.71 mL

Explanation:

Step 1: Data given

Volume of water =150 mL = 0.150 L

concentration of sucrose solution 35 % w/w this means in 100 grams of water we have 35 grams of sucrose

specific gravity =1.115

Step 2: Calculate the density of the solution

Density = specific gravity * density of water

Density of solution = 1.115 * 1g/ mL

Density of solution = 1.115 g/ mL

Step 3: Calculate mass of the solution

Mass of solution = density ¨volume

Mass of solution = 1.115 g/ mL * 150 mL

Mass of solution = 167.25 grams

Step 4: Calculate mass of sucrose

35 % = 0.35 * 167.25 grams

Mass sucrose = 58.54 grams

Step 5: Calculate mass of water

Mass of water = mass of sample - mass of sucrose

Mass of water = 167.25 - 58.54 = 108.71 grams

Step 6: Calculate volume of water

Volume = mass / density

Volume = 108.71 grams / 1g/ mL

Volume = 108.71 mL = 0.10871 L

The volume of the water is 108.71 mL

Answer 2

Answer:

V H2O = 108.713 mL

Explanation:

mix: H2O + C12H22O11

∴ Vmix = 150 mL

∴ %m/m C12H22O11 = 35% = (m C12H22O11 / m mix)×100

∴ SG = 1.115 = ρ mix / ρ H2O

∴ ρ H2O = 1 g/mL

⇒ ρ mix = 1.115 g/mL = m mix/ V mix

⇒ m mix = (1.115 g/mL)×(150 mL) = 167.25 g

⇒ V H2O = ?

⇒ 0.35 = m C12H22O11 / m mix

⇒ (0.35)×(m mix) = m C12H22O11

⇒ m C12H22O11 = (0.35)×(167.25 g) = 58.54 g

⇒ m mix = m H2O + m C12H22O11

⇒ m H2O = 167.25 - 58.54 = 108.7125 g

∴ ρ H2O = m H2O / V H2O

⇒ V H2O = (108.7125 g / 1 g/mL) = 108.7125 mL