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meriva
3 years ago
15

Two unknown molecular compounds were being studied. A solution containing 5.00 g of compound A in 100. g of water froze at a low

er temperature than a solution containing 5.00 g of compound B in 100. g of water. Which compound has the greater molar mass
Chemistry
1 answer:
LenaWriter [7]3 years ago
4 0

Answer:

Compound B has greater molar mass.

Explanation:

The depression in freezing point is given by ;

\Delta T_f=i\times k_f\times m..[1]

m=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Mass of solvent in kg}}

Where:

i = van't Hoff factor

k_f = Molal depression constant

m = molality of the solution

According to question , solution with 5.00 g of A in 100.0 grams of water froze at at lower temperature than solution with 5.00 g of B in 100.0 grams of water.

The depression in freezing point of solution with A solute: \Delta T_{f,A}

Molar mass of A = M_A

The depression in freezing point of solution with B solute: \Delta T_{f,B}

Molar mass of B = M_B

\Delta T_{f,A}>\Delta T_{f,B}

As we can see in [1] , that depression in freezing point is inversely related to molar mass of the solute.

\Delta T_f\propto \frac{1}{\text{Molar mass of solute}}

M_A

This means compound B has greater molar mass than compound A,

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It indicates the number of moles of reactants and products

Explanation:

The coefficients in front of the reactants and products in a chemical reaction represents the number of moles of reactants and products.

Every reaction is made up of equal number of moles of reactants and products. Thus, chemical equations are written in such a way to obey the law of conservation of matter.

The numbers used are usually whole numbers and the are very important in stoichiometry.

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How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?
Ksenya-84 [330]

Answer:

a. Approximately 1.3\; \rm mL.

b. Approximately 7.2\; \rm mL.

Explanation:

The unit of concentration "\rm M" is equivalent to "\rm mol \cdot L^{-1}", which means "moles per liter."

However, the volume of both solutions were given in mililiters \rm mL. Convert these volumes to liters:

\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L.

\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L.

In a solution of volume V where the concentration of a solute is c, there would be c \cdot V (moles of) formula units of this solute.

Calculate the number of moles of \rm NaCl formula units in each of the two solutions:

Solution in a.:

n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol.

Solution in b.:

n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol.

What volume of that 3.4\; \rm M (same as 3.4 \; \rm mol \cdot L^{-1}) \rm NaCl solution would contain that many

For the solution in a.:

\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL.

Similarly, for the solution in b.:

\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL.

8 0
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Answer:

7

Explanation:

The mass number is neutrons plus protons

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