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ahrayia [7]
3 years ago
14

1.2 g of sucrose is added to water with the final volume of 75 ml. What is the concentration of sucrose in %w/v?

Chemistry
1 answer:
Firlakuza [10]3 years ago
8 0

Answer:

16 g/L

Explanation:

Given that:-

Mass of sucrose added = 1.2 g

Volume of water = 75 mL = 0.075 L ( 1mL = 0.001 L )

%w/v is defined as the mass of the solution in 1 L of the solution.

So,

\%\ w/v=\frac{Mass_{solute}}{Volume_{Solution}}

\%\ w/v=\frac{1.2\ g}{0.075\ L}=16\ g/L

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<span>Since,
1000 grams of water = 1000 mL of water</span><span>
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</span>1000 mL = 10 x 100 mL
<span>
moles of NH4Cl = 53.5/53.49
                          = 1.0 m
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A different student is given a 10.0g sample labeled CaBr2 that may contain an inert (nonreacting) impurity. Identify a quantity
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Answer:

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Explanation:

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8 0
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3 years ago
A 720. cm^3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.966 g at 25.0°C and the pressure is 760.
sleet_krkn [62]

Explanation:

The given data is as follows.

      Pressure (P) = 760 torr = 1 atm

      Volume (V) = 720 cm^{3} = 0.720 L

     Temperature (T) = 25^{o}C = (25 + 273) K = 298 K

Using ideal gas equation, we will calculate the number of moles as follows.

                                PV = nRT

   Total atoms present (n) = \frac{PV}{RT}

                                          = 1 \times \frac{0.720 L}{0.0821 \times 298}

                                           = 0.0294 mol

Let us assume that there are x mol of Ar and y mol of Xe.

Hence, total number of moles will be as follows.

               x + y = 0.0294

Also,      40x + 131y = 2.966

             x = 0.0097 mol

              y = (0.0294 - 0.0097)

                = 0.0197 mol

Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = \frac{y}{(x+y)}

                               = \frac{0.0197}{0.0294}

                              = 0.67

Therefore, the mole fraction of Xe is 0.67.

6 0
2 years ago
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