Suppose now that you wanted to determine the density of a sm

Suppose now that you wanted to determine the density of a small yellow crystal to confirm that it is silicon. From the literature, you know that silicon has a density of 2.33 g/cm³. How would you prepare 20.0 mL of the liquid mixture having that density from pure samples of  CHCl₃ (d = 1.492 g/mL) and  CHBr₃ (d = 2.890 g/mL)? (Note: 1 mL = 1 cm³.)

V(CHCl₃) =  mL

V(CHBr₃) =  mL

Solution

Ans. #Step 1: Let the required volume of CHCl₃ = X mL

            And, required volume of CHBr₃ = Y mL

So, sum of volumes must be equal to the total volume of solution required.

Or,       X mL + Y mL = 20.0 mL

            Hence, X + Y = 20.0                        - equation

# Step 2: The required density of solution mixture is assumed to be 2.33 g/ cm³ to test if the silicon crystal’s density equal to it.

So, mass of liquid mixture = Volume x Density

                                                = 20.0 mL x (2.33 g/ mL)

                                                = 46.6 g

# Mass of X mL CHCl₃ = X mL x (1.492 g/ mL) = 1.492X g

Mass of Y mL CHBr₃ = Y mL x (2.890 g/ mL) = 2.890Y g

Now, sum of masses of the two components must be equal to the mass of 20.0 mL liquid mixture.

So,

            1.492X g + 2.890Y g = 46.6 g

            Hence, 1.492X + 2.890Y = 46.6               - equation 2

# Step 3: Comparing (equation 1 x 1.492) – equation 2

                 1.492X + 1.492Y = 29.84

            (-) 1.492X + 2.890Y = 46.60

            ------------------------------

                               -1.398Y = -16.76

                        Or, Y = 16.76 / 1.398 = 11.9886

Therefore, required volume of CHBr₃ = Y mL = 11.9886 mL

# Putting the value of Y in equation 1-

            X = 20.0 mL – 11.9886 mL = 8.0114 mL

Hence, required volume of CHCl₃ = 8.0114 mL