Question

The density of a 1.0 M solution of NaCl is 1.04 g/mL and its specific heat, c, is 3.06 J/C*g. How much heat is absorbed by 3.14 mL of this solution if the temperature rises from 18.6C to 29.7C?

q=m*C*delta T

(1.04g)(3.06)(11.1K)=35.32 is that right?

Answer 1

1.04(g/ml)*3.06(J/g/°K)*11.1(°K)
The expression is missing the volume of 3.14 ml.

However, I have doubts as to the value of 3.06 J/g/°K which seems low.
Was there a typo? 4.06 seems to be in the right order of magnitude.