The Salinity (Mostly NaCl) in Some Parts of the Great$27 \%$

Question

Examlex · Accepted Answer

The Answer is From Example 14.4, we may write

d = \frac { i R M T } { ho g }

where, for salt water

i = 2

and

R = 8.315 \mathrm {~J} \cdot \mathrm { K } ^ { - 1 } \cdot \mathrm { mol } ^ { - 1 }

. For

T = 283 \mathrm {~K} , ho = 1200 \mathrm {~kg} / \mathrm { m } ^ { 3 }

and

g

= 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 }

, then

d = \left[ \frac { 2 imes \left( 8.315 \mathrm {~J} \cdot \mathrm { K } ^ { - 1 } \cdot \mathrm { mol } ^ { - 1 } ight) imes ( 283 \mathrm {~K} ) } { \left( 1200 \mathrm {~kg} / \mathrm { m } ^ { 3 } ight) imes \left( 9.8 \mathrm {~m} / \mathrm { s } ^ { 2 } ight) } ight] M = \left[ 0.40 \frac { \mathrm { m } ^ { 4 } } { \mathrm {~mol} } ight] M

where

M

is the molarity in

\mathrm { mol } / \mathrm { m } ^ { 3 }

and

d

is given in

\mathrm { m }

. The molecular mass of

\mathrm { NaCl }

is

58.4

\mathrm { g } / \mathrm { mol }

so that

1 \mathrm {~mol} / \mathrm { L }

is

58.4 \mathrm {~g} / \mathrm { kg }

or

5.84 \% \mathrm { NaCl } \mathrm { in } ^ { 2 } \mathrm { H } _ { 2 }

by weight. So

M = ( \mathrm { wt } \% ) imes \left( 1000 \mathrm {~mol} / \mathrm { m } ^ { 3 } ight) / ( 5.84 \mathrm { wt } \% ) = \left( 171.2 \mathrm {~mol} / \left( \mathrm { m } ^ { 3 } \mathrm { wt } \% ight) ight) imes \mathrm { wt } \%

Then for

27 \% , M = \left( 171.2 \mathrm {~mol} / \left( \mathrm { m } ^ { 3 } \mathrm { wt } \% ight) ight) imes ( 27 \mathrm { wt } \% ) = 4623 \mathrm {~mol} / \mathrm { m } ^ { 3 }

and

d = \left( 0.40 \mathrm {~m} ^ { 4 } / \mathrm { mol } ight) imes \left( 4623 \mathrm {~mol} / \mathrm { m } ^ { 3 } ight) = 1849 \mathrm {~m}

The Salinity (Mostly NaCl) in Some Parts of the Great $27 \%$