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3 years ago

How many moles are represented in 213 grams of sodium bromide , NaBr?

Chemistry

1 answer:

Eduardwww [97]3 years ago

5 0

Answer:

The number of mole represented is 2.07 moles

Explanation:

Using the formula

n = m/Mm

n - mole

m - mass

Mm- molar mass

Since the mass is given to be 213g

So lets calculate the molar mass of sodium bromide (NaBr)

The molar mass of Na = 22.99

Br = 79.904

Molar mass = 22.99+79.904

= 102.894g/mol

Using the formula n = m/Mm

n = 213g/ 102.894g/mol

n = 2.07moles

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Putting values in above equation, we get:

$1.034g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.034g/mL\times 1000mL)=1034g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of glucose = 0.495 moles

Molar mass of glucose = 180.16 g/mol

Putting values in above equation, we get:

$0.495mol=\frac{\text{Mass of glucose}}{180.16g/mol}\\\\\text{Mass of glucose}=(0.495mol\times 180.16g/mol)=89.18g$

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

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$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

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Freezing point of pure solution = 0°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 1.86°C/m

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