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20 October, 17:01

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

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Answers (2)
  1. 20 October, 17:55
    0
    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
  2. 20 October, 20:10
    0
    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
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