The grams of 22.9 % sugar solution that contain 68.5 g of sugar is 299.13 g of solution
<u><em>calculation</em></u>
22.9% means that there are 22.9 g of sugar in 100 g of solution.
what about 68.5 g of sugar
- <em>by cross multiplication</em>
=[(68.5 g sugar x 100 g solution) /22.9 g sugar] =299.13 g of solution
Nb; <em>g sugar cancel each other</em>
Answer:
the Ray's go right through it
Explanation:
the Ray's are so small they punch right through them
A solution (in this experiment solution of NaNO₃) freezes at a lower temperature than does the pure solvent (deionized water). The higher the
solute concentration (sodium nitrate), freezing point depression of the solution will be greater.
Equation describing the change in freezing point:
ΔT = Kf · b · i.
ΔT - temperature change from pure solvent to solution.
Kf - the molal freezing point depression constant.
b - molality (moles of solute per kilogram of solvent).
i - Van’t Hoff Factor.
First measure freezing point of pure solvent (deionized water). Than make solutions of NaNO₃ with different molality and measure separately their freezing points. Use equation to calculate Kf.
6 - one sodium atom, 1 hydrogen atom, 1 carbon atom, and 3 oxygen atoms.
Answer:
In 1935
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