## General Chemistry

### Chemistry of Solutions

**7 out of 10**

The concentration of solutions are generally expressed in terms of molarity, molality, normality, and weight percent. The formation of the solutions itself has many ramifications. The solubility of solutes differ considerably from one other. Some of the factors that can influence the solubility include temperature and pressure. Solubility depends on other factors as well. Given below in Figure-1 is a graph which depicts solubility differences of some solutes.

Quite often, the freezing and boiling point changes that are brought about by the dissolved solutes can be predicted reasonably. But this is not always the case.

The predictions and calculations are done for freezing point depression (ΔT_{f}) based on the following formula:

ΔT_{f} = K_{f} m,

where m is the molality and K_{f} is the freezing point depression constant. (K_{f}= 1.86 ^{o}C/*m*)

For boiling point elevation (ΔT_{b}), the calculations are based on the following formula:

ΔT_{b} = K_{b} m,

where m is the molality and K_{b} is the boiling point elevation constant. (K_{b} = 0.512 ^{o}C/*m*)

A solution was made by using 315 g of glucose in 750 g of water. What is the boiling point of this solution?

A | 98.8^{0} C |

B | 99.8^{0} C |

C | 100.24^{0} C |

D | 101.2^{0} C |

**Ans. D**The first step in answering this question is to find the molality of the solution. Then plug that number into the boiling-point-change equation given in the passage.

ΔT_{b} = K_{b} m = (0.512^{0} C/m)(2.33 m) = 1.2^{0} C.

The answer is 100^{0} C + 1.2^{0} C = 101.2^{0} C.

##
Chemistry of Solutions Flashcard List

**10**flashcards

1) |

2) |

3) |

4) |

5) |

6) |

7) |

8) |

9) |

10) |