Chemistry, 21.07.2019 06:30 tyme66

What is the melting point of a solution in which 3.5 grams of sodium chloride is added to 230 ml of water?

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We are going to use this equation:

ΔT = - i m Kf

when m is the molality of a solution

i = 2

and ΔT is the change in melting point = T2- 0 °C

and Kf is cryoscopic constant = 1.86C/m

now we need to calculate the molality so we have to get the moles of NaCl first:

moles of NaCl = mass / molar mass

= 3.5 g / 58.44

= 0.0599 moles

when the density of water = 1 g / mL and the volume =230 L

∴ the mass of water = 1 g * 230 mL = 230 g = 0.23Kg

now we can get the molality = moles NaCl / Kg water

=0.0599moles/0.23Kg

= 0.26 m

∴T2-0 = - 2 * 0.26 *1.86

∴T2 = -0.967 °C

Answer from: ansley81

0.952 °C

Explanation:

The change in melting point is computed as:

ΔT = k*m*i

where ΔT is the difference between the melting point of water and of solution, k is a constant (1.86 °C*kg/mol for water), i is the van't Hoff factor (equal to 2 for sodium chloride because 2 ions are obtained after its dissolution), and m is the molality of the solution.

Molar mass of sodium chloride: 58.44 g/mol

Moles of of sodium chloride: mass / molar mass 3.5/58.44 = 0.059 mol

Density of water 1 kg/L

230 mL of water are equivalent to 0.23 L

mass of water: density * volume = 1*0.23 = 0.23 kg

Molality of the solution: m = moles of solute/ kg of solvent = 0.059/0.23 = 0.256

Finally:

ΔT = 1.86*0.256*2 = 0.952 °C

Water melting point: 0 °C

So, the solution melting point is: 0 - 0.952 = 0.952 °C

Answer from: Quest

29.1-12=17.1the answer is this one

Answer from: Quest

each column of the table is called a group. elements in a group share similar properties. groups are read from top to bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o. 2. 3. 4. 5. 6. 7. 8. 9. 2. beryllium. 9.012. 4. sodium. 22.990. 11. 3. magnesium. 24.305. 12. potassium. 39.098. 19. 4. calcium. 40.078. 20. i guess i

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What is the melting point of a solution in which 3.5 grams of sodium chloride is added to 230 ml of...

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