Question

What is the boiling point of an aqueous solution containing 117 g NaCl in 1000 g $H_{2}O$? $K_b={0.52}^{o}Ckg/mol$

• A. ${98.96}^{o}C$
• B. ${99.48}^{o}C$
• C. ${100.52}^{o}C$
• D. ${101.04}^{o}C$
• E. ${102.08}^{o}C$

Answer 1

The correct option is E ${102.08}^{o}C$

$Molarity$$=$$117/(58.5\times 1)$$=$$2M$

So change in temperature$=$$i\times k_b\times molarity$

or change in temperature$=$$2\times 0.52\times 2$$=$$2.08$

hence boiling point$=$$100+2.08$$=$$102.08$