Question

The compressibility of water is $46.4\times 10^{-6} atm^{-1}$. This means that:
A. the bulk modulus of water is $46.4\times 10^{6} atm^{-1}$
B. the volume of water decreases by $46.4\times$ one-millionths of the original volume for each atmosphere increase in pressure
C. when water is subjected to an additional pressure of one atmosphere its volume decreases by $46.4\%$
D. when water is subjected to an additional pressure of one atmosphere its volume is reduced to $10^{-6}$ of its original volume.

Answer 1

Clearly, option C and D is not the answer. As in C, if we convert $46.4\times 10^{-6} atm^{-1}$ to percentage, we get $46.4\times 10^{-8}\%$. And in D, the value $46.4$ is neglected. Leaving behind option A and B. To find the correct option, let use the definition of compressibility.

Formula used:
$K=\dfrac{\Delta P} {\dfrac{\Delta V}{V}}$ and $C=\dfrac{1}{K}$

Complete step by step answer:
We know that, compressibility is defined as the reciprocal of bulk modulus, and bulk modulus is defined as the ratio of pressure applied to the change in volume.
Given that, compressibility $C$ of the water is $46.4\times 10^{-6} atm^{-1}$, then $C=\dfrac{1}{K}$. Then bulk modulus $K$ is $46.4\times 10^{6} atm^{1}$. But here A. is given as the bulk modulus of water is $46.4\times 10^{6} atm^{-1}$. See that though the value is correct, the unit is wrong. Thus it is not the correct option. Also the unit of bulk modulus is $N/m^{2}$ and not $atm$.
Then the only option left is B. Let us check it also.
As we know that $K=\dfrac{\Delta P} {\dfrac{\Delta V}{V}}$ from its definition. Then rearranging, we get $\dfrac{\Delta P}{K}=\dfrac{\Delta V}{V}$. If $\Delta P=1$, we have $46.4\times 10^{-6} =\dfrac{\Delta V}{V}$.
Where $\dfrac{\Delta V}{V}$, is the fractional change in the volume.
Clearly, this implies that the volume of water decreases by $46.4\times$ one-millionths of the original volume for each atmosphere increase in pressure.
Thus the answer is B. the volume of water decreases by $46.4\times$ one-millionths of the original volume for each atmosphere increase in pressure.

Note:
To find the correct option start discarding options. Then use the definition of the quantity to identify the options. Also be careful with the missing terms, or powers and units. Here though all the options look similar, there are small changes in the units and terms.