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(c) The product sodium chloride is dissolved in water to separate it from titanium. At 30^circ C the solubility of sodium chloride is 36 kg per 100dm^3 Calculate the minimum volume of water in dm^3 at 30^circ C needed to dissolve 1989 kg sodium chloride. __ Volume of water =underline ( )dm^3

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(c) The product sodium chloride is dissolved in water to separate it from titanium.
At 30^circ C the solubility of sodium chloride is 36 kg per 100dm^3
Calculate the minimum volume of water in dm^3 at 30^circ C needed to dissolve 1989 kg
sodium chloride.
__
Volume of water =underline ( )dm^3

(c) The product sodium chloride is dissolved in water to separate it from titanium. At 30^circ C the solubility of sodium chloride is 36 kg per 100dm^3 Calculate the minimum volume of water in dm^3 at 30^circ C needed to dissolve 1989 kg sodium chloride. __ Volume of water =underline ( )dm^3

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KristinaMaster · Tutor for 5 years

Answer

After performing the calculation, we find that \(x = 5525 \mathrm{dm}^{3}\). Therefore, the minimum volume of water needed to dissolve \(1989 \mathrm{~kg}\) of sodium chloride at \(30^{\circ} \mathrm{C}\) is \(5525 \mathrm{dm}^{3}\).

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## Step 1:<br />First, we need to understand the problem. The solubility of sodium chloride at \(30^{\circ} \mathrm{C}\) is given as \(36 \mathrm{~kg}\) per \(100 \mathrm{dm}^{3}\) of water. This means that to dissolve \(36 \mathrm{~kg}\) of sodium chloride, we need \(100 \mathrm{dm}^{3}\) of water. We are asked to find the minimum volume of water needed to dissolve \(1989 \mathrm{~kg}\) of sodium chloride.<br /><br />## Step 2:<br />We can set up a proportion to find the volume of water needed to dissolve \(1989 \mathrm{~kg}\) of sodium chloride. The proportion is set up as follows:<br /><br />### \(\frac{36 \mathrm{~kg} \mathrm{NaCl}}{100 \mathrm{dm}^{3} \mathrm{H_{2}O}} = \frac{1989 \mathrm{~kg} \mathrm{NaCl}}{x \mathrm{dm}^{3} \mathrm{H_{2}O}} \)<br /><br />## Step 3:<br />Next, we solve the proportion for \(x\), which represents the volume of water needed to dissolve \(1989 \mathrm{~kg}\) of sodium chloride.<br /><br />### \(x = \frac{1989 \mathrm{~kg} \mathrm{NaCl} \times 100 \mathrm{dm}^{3} \mathrm{H_{2}O}}{36 \mathrm{~kg} \mathrm{NaCl}} \)
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