Respuesta :

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[tex]d_{c}=\frac{m_{1}+m_{2}}{V_{1}+V_{2}}=\frac{10mL*1,492\frac{g}{mL}+5mL*2,89\frac{g}{mL}}{15mL}=\\ \\ =\frac{14,92g+14,45g}{15mL}=\frac{29,37g}{15mL}=1,958\frac{g}{mL}[/tex]

Answer:

1.958 g/mL

Explanation:

the density of mixture will be calculated by taking ratio of total mass of the mixture obtained by mixing the two liquids and the total volume of the mixture

the total volume = volume of CHCl₃ + volume of  CHBr₃ = 10.0+5.0 =15.0mL

the mass of CHCl₃=density of CHCl₃ X volume of CHCl₃

                               = 1.492 X 10.0  = 14.92 g

the mass of CHBr₃=density of CHBr₃ X volume of CHBrl₃

                               = 2.890 X 5.0  = 14.45 g

Thus total mass = 14.92+14.45 = 29.37

Density of mixture = total mass / total volume = 29.37 / 15.0 = 1.958g/mL

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