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Chemistry

What weight of sodium nitrate will separate when a saturated solution containing 50 gram of water is cooled from 50°C to 30°C ? The solubility of NaNO₃ at 50°C and 30°C is 114 g and 86 g respectively.

Water

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Answer

Given,
Mass of solvent = 50 g

Solubility of NaNO₃ at 50°C = 114 g

Solubility of NaNO₃ at 30°C = 86 g

Solubility = $\dfrac{\text{Mass of solute}}{\text{Mass of solvent}}$ x 100

Rearranging the above equation

Mass of solute = $\dfrac{\text{Solubility x Mass of solvent}}{100}$

∴ Mass of solute at 50°C = $\dfrac{114}{100}$ x 50

= 57 g

∴ Mass of solute at 30°C = $\dfrac{86}{100}$ x 50

= 43 g

∴ Amount of NaNO₃ that separate = 57 g - 43 g = 14 g

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Related Questions

The following table gives the solubility of different salts at different temperatures.
Temperature (in K) Substance dissolved (in g)
KNO₃ NaCl KCl NH₄Cl
283 21 36 35 24
293 32 36 35 37
313 62 36 40 41
333 106 37 46 55
353 167 37 54 66
Answer the following questions based on the table given above.
(a) What mass of KNO₃ would be needed to produce a saturated solution of KNO₃ in 50 grams of water at 313 K.
(b) If a saturated solution of KCl is made at 353 K and then cooled at room temperature, what would you observe? Explain.
(c) Find the solubility of each salt at 293 K.
(d) Which salt has the lowest solubility at 283 K?
(e) What is the effect of change of temperature on the solubility of a salt.
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