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Shkiper50 [21]
4 years ago
6

When 1.47 g of an unknown non-electrolyte is dissolved in 50.0 g of carbon tetrachloride, the freezing point decreased by 7.03 d

egrees C. If the Kfp of the solvent is 29.8 K/m, calculate the molar mass of the unknown solute.
Chemistry
1 answer:
CaHeK987 [17]4 years ago
3 0

Answer:

124.6 g/mol will be the molar mass for the unknown non-electrolyte

Explanation:

Freezing point depression →  ΔT = Kf . m

ΔT indicates the temperature variation:

Freezing point of pure solvent - Freezing point of solution

Kf is the cryoscopic constant → 29.8 K/m

Generally the unit is °C/m but we have to type the ΔT by K so let's replace the data given → 7.03 K = 29.8 K/m . m

m = 7.03 K / 29.8 m/K = 0.236 mol/kg

m means molality, the moles of solute in 1 kg of solvent. If we want to determine the moles of solute we do this operation:

molality . kg, so we may convert the mass of solvent from g to kg.

50 g . 1kg/1000g = 0.05 kg

0.236 mol/kg . 0.05 kg = 0.0118 moles

Molar mass → g/mol → So 1.47 g / 0.0118 mol = 124.6 g/mol

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A sample 0.100 moles of a gas is collected at at stp. what is the volume of the gas in liters? group of answer choices 2.44 l
bearhunter [10]

n = 0.100 moles

At STP (standard temperature pressure), where gas is collected, the temperature is 0°C and the pressure is 1 atm.

\\$\therefore$ Temperature $(T)=0^{\circ} \mathrm{C}=0^{\circ}+273

                              =273K

          Pressure $(P)=1$ atm.

The ideal gas equation indicates that

\begin{aligned}& P V=n R T \Rightarrow V=\frac{n R T}{P} \\\therefore & R=\text { gas constant }=0.0821 \mathrm{~L} \cdot \mathrm{cetm} / \mathrm{mo} / \cdot K\end{aligned}

 \begin{aligned}&V=\frac{0.100 \mathrm{~mol} \cdot \times 0.0821 \mathrm{~L} \cdot \mathrm{atm} / \mathrm{mol} . \mathrm{K} \times 273 \mathrm{~K}}{1 \mathrm{~atm}} \\&V=\frac{2.24 L}{1}\end{aligned}\\

V=2.24L

Group of answer choices 2.44L .The volume occupied by  mole of a given gas at a given temperature and pressure is expressed as the gas's molar volume.

The most typical illustration is the molar volume of a gas at STP, which is equal to 2.44L for 1 mole of any ideal gas at a temperature of 273.15 Kand a pressure of 1 atm.

Hence, the volume of the gas is 2.24L.

<h3 /><h3>What is the STP?</h3>

A unit is stated to have a temperature of absolute zero (273 Kelvins) and an atmospheric pressure of one atmosphere, or 1 atm, at standard temperature and pressure. Additionally, at STP, a mole of any gas takes up 22.414 L of space. Keep in mind that this idea only applies to gases.

For experimental measurements to be established under standard conditions that allow for comparisons between various sets of data, standard temperature and pressure must be met.

To learn more about STP, Visit:

brainly.com/question/1626157

#SPJ4

8 0
2 years ago
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