QuestionJuly 15, 2025

A container has 9.0 ml of liquid X with a density of 1200kg/m^3 How much liquid Y with a density of 1400kg/m^3 should be added to the flask so that the density of the liquid mixture is 1280kg/m^3 3.5 ml 6.0 ml 6.4 ml 8.7 ml

A container has 9.0 ml of liquid X with a density of 1200kg/m^3 How much liquid Y with a density of 1400kg/m^3 should be added to the flask so that the density of the liquid mixture is 1280kg/m^3 3.5 ml 6.0 ml 6.4 ml 8.7 ml
A container has 9.0 ml of liquid X with a density of 1200kg/m^3 How much liquid Y with a density of 1400kg/m^3 should be added to the flask so that the density of the liquid mixture is 1280kg/m^3 3.5 ml 6.0 ml 6.4 ml 8.7 ml

Solution
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Answer

6.0 ml Explanation 1. Calculate mass of liquid X Mass of liquid X = Volume × Density = 9.0 \, \text{ml} \times 1200 \, \text{kg/m}^{3}. Convert ml to m³: 9.0 \, \text{ml} = 9.0 \times 10^{-6} \, \text{m}^{3}. Mass = 9.0 \times 10^{-6} \, \text{m}^{3} \times 1200 \, \text{kg/m}^{3} = 0.0108 \, \text{kg}. 2. Set up equation for mixture density Let volume of liquid Y be V_Y ml. Total mass = Mass of X + Mass of Y = 0.0108 \, \text{kg} + V_Y \times 10^{-6} \, \text{m}^{3} \times 1400 \, \text{kg/m}^{3}. Total volume = (9.0 + V_Y) \times 10^{-6} \, \text{m}^{3}. Mixture density = \frac{\text{Total mass}}{\text{Total volume}} = 1280 \, \text{kg/m}^{3}. 3. Solve for V_Y \frac{0.0108 + V_Y \times 10^{-6} \times 1400}{(9.0 + V_Y) \times 10^{-6}} = 1280. Simplify and solve: 0.0108 + 1400 \times V_Y \times 10^{-6} = 1280 \times (9.0 + V_Y) \times 10^{-6}. Solve for V_Y: V_Y = 6.0 \, \text{ml}.

Explanation

1. Calculate mass of liquid X<br /> Mass of liquid X = Volume × Density = $9.0 \, \text{ml} \times 1200 \, \text{kg/m}^{3}$. Convert ml to m³: $9.0 \, \text{ml} = 9.0 \times 10^{-6} \, \text{m}^{3}$. Mass = $9.0 \times 10^{-6} \, \text{m}^{3} \times 1200 \, \text{kg/m}^{3} = 0.0108 \, \text{kg}$.<br />2. Set up equation for mixture density<br /> Let volume of liquid Y be $V_Y$ ml. Total mass = Mass of X + Mass of Y = $0.0108 \, \text{kg} + V_Y \times 10^{-6} \, \text{m}^{3} \times 1400 \, \text{kg/m}^{3}$. Total volume = $(9.0 + V_Y) \times 10^{-6} \, \text{m}^{3}$. Mixture density = $\frac{\text{Total mass}}{\text{Total volume}} = 1280 \, \text{kg/m}^{3}$.<br />3. Solve for $V_Y$<br /> $\frac{0.0108 + V_Y \times 10^{-6} \times 1400}{(9.0 + V_Y) \times 10^{-6}} = 1280$. Simplify and solve: $0.0108 + 1400 \times V_Y \times 10^{-6} = 1280 \times (9.0 + V_Y) \times 10^{-6}$. Solve for $V_Y$: $V_Y = 6.0 \, \text{ml}$.
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