Chapter 2, Problem 66A

Interpretation Introduction

Interpretation: The solid that melt before mercury boils needs to be identified when an increase in temperature takes place.

Concept Introduction: The melting point is defined as the temperature at which a solid substance starts melting. The boiling point is defined as the temperature of a liquid when its vapor pressure is equal to the atmospheric pressure. This is the temperature when liquid starts boiling and converts into vapor.

Answer to Problem 66A

Sulphur.

Explanation of Solution

On checking the temperature ranges, bromine will be liquid in the temperature range $-{\text{7 }}^{\text{o}}{\text{C to 59 }}^{\text{o}}\text{C }$ , chlorine will be gas above ${\text{-34 }}^{\text{o}}\text{C}$ , ethanol is liquid in the temperature range $-11{\text{7 }}^{\text{o}}{\text{C to 78 }}^{\text{o}}\text{C }$ , mercury is liquid in the temperature range $-39{}^{\text{o}}{\text{C to 357 }}^{\text{o}}\text{C }$ , neon is gas above $-{\text{246 }}^{\text{o}}\text{C}$ , water is liquid in the range $0{}^{\text{o}}\text{C}-100{}^{\text{o}}\text{C}$ and sulphur is solid in the range $115{}^{\text{o}}\text{C}-445{}^{\text{o}}\text{C}$ . Here, only sulphur is solid at room temperature.

Thus, the boiling point of mercury is ${\text{357 }}^{\text{o}}\text{C}$ now, the solid that melts before mercury boils will have melting point below ${\text{357 }}^{\text{o}}\text{C}$ . Thus, the correct answer is sulphur.

Conclusion

Thus, sulphur is solid that melt before mercury boils.