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Ahat [919]
3 years ago
9

What volume of a 2.10 M NaCl solution is

Chemistry
1 answer:
cluponka [151]3 years ago
3 0

Answer:

146.3g NaCl (mol NaCl/58.44g NaCl) = 2.50 mol NaCl

1.5M NaCl = 1.5 mol NaCl / 1 L = 2.5 mol NaCl / x L, solve for x

x L = 2.5 mol NaCl / 1.5 mol NaCl = 1.66 L

It gives the answer and all the working.

To put it another way:

Dividing the amount required by the molar mass

we quickly see that 2.5 moles are required.

One litre of 1.5 molar solution gives 1.5 moles

we need a further mole, which is 2/3 of 1.5 so 2/3 of a litre.

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If you added 45,000 calories to water that was at 25 degrees C, and the ending temperature was 35 degrees C, how much water did
user100 [1]

<u>Answer:</u>

<em>4.5 L water we have in litres (L).</em>

<em><u></u></em>

<u>Explanation:</u>

Q=m\times c \times \Delta T

where

\Delta T = Final T - Initial T

Q is the heat energy in calories

c is the specific heat capacity (for water 1.0  cal/(g℃))  

m is the mass of water

Plugging in the values  

\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times\left(35^{\circ} \mathrm{C}-25^{\circ} \mathrm{C}\right)$\\\\$45000 \mathrm{cal}=m \times 1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}$\\\\$m=\frac{45000 \mathrm{cal}}{1.0 \frac{\mathrm{cal}}{\mathrm{g}^{\circ} \mathrm{C}} \times 10^{\circ} \mathrm{C}}$\\\\$m=4500 \mathrm{g}$\\\\Density of water $=\frac{\text { mass }}{\text { volume }}$

So,

Volume of water = mass/density

\\\\=\frac{4500 \mathrm{g}}{\frac{1.09}{\mathrm{mL}}}=4500 \mathrm{mL}$$

=4.5 L (Answer)

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