Question

Calculate the molality of solution prepared by dissolving 50g NaOH in 100g water?​

Answer 1

ANAWER: 12.5M

EXPLANATION -

What is the molarity of the solution prepared by dissolving 50g of NaOH in 100ml of water?I’m sure others will answer, but … you should be specific whether you are referring to 0.1 L as the volume of pure water, or the volume of the resultant solution. These are not the same, due to the fact that in most solutions, volumes are not strictly additive, as there is a volume difference due to the interaction of solute (in this case anhydrous NaOH) and solvent (pure water), the so-called delta V of mixing (itself a function of concentration). If I assume that you mean 0.1L solution, then the answer is simple, and I’ll leave it to others to answer. Otherwise …Molarity is moles per volumeSounds like a homework problem. Since molarity (M) is defined as moles of solute per L of solution. Therefore, convert the given mass of NaOH into moles of NaOH by dividing the 50 grams of INaOH by the molecular mass of NaOH which is approximately 50 g /(40 g/mol) or 1.25 mol NaOH. Then divide the moles of NaOH by 0.100 L water. Note: you actually need to divide by L of solution which accounts for both the volume of solute (NaOH) and solvent (water). However, you dont know if the volume of the solid NaOH is additive to the 100 mL of water. Assuming the NaOH volume has a density of 2.13 g/mL then it’s volume is 50 g /(2.13 g/mL) = 23.474 mL or 0.0235 L. If we assume the actual solution volume of solute is additive to the 100 mL of water then the Molarity is equal to 1.25 mol/0.1235 L or 10.12 M. However, if we assume the NaOH volume is not additive then the Molarity of the solution is 1.25 mol/0.100 L or 12.5 M.

MARK BRAINLIEST

Answer 2

Answer:

Explanation:

Concept:

The amount of a material dissolved in a given mass of solvent is known as molality (m) or molal concentration. It's the number of moles of a solute per kilogramme of solvent.

Given:

Weight of $NaOH$ is $50g$.

Weight of $H_{2} O$ is $100g$.

Find:

Determine the molality of the solution.

Solution:

Let's consider the formula of molality

$molality =\frac{moles of solute}{kilograms of solvent}$

Now insert the values provided,

Moles of $NaOH$$=\frac{Weight of NaOH}{Mass of NaOH}$

                           $=\frac{50}{40} \\=\frac{5}{4}\\=1.25$

kilograms of $H_{2} O$ $=\frac{Weight of H_{2} O}{1000}$

                            $\ =\frac{100}{1000} \\=0.1$

Hence, molality$= \frac{moles of solute}{kilograms of Solvent}$

$= \frac{1.25}{0.1} \\=12.5m$

The molality of solution is $12.5m.$

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