Answer : The mass of sodium bromide added should be, 18.3 grams.
Explanation :
Molality : It is defined as the number of moles of solute present in kilograms of solvent.
Formula used :
![Molality=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Mass of solvent}}](https://tex.z-dn.net/?f=Molality%3D%5Cfrac%7B%5Ctext%7BMass%20of%20solute%7D%5Ctimes%201000%7D%7B%5Ctext%7BMolar%20mass%20of%20solute%7D%5Ctimes%20%5Ctext%7BMass%20of%20solvent%7D%7D)
Solute is, NaBr and solvent is, water.
Given:
Molality of NaBr = 0.565 mol/kg
Molar mass of NaBr = 103 g/mole
Mass of water = 315 g
Now put all the given values in the above formula, we get:
![0.565mol/kg=\frac{\text{Mass of NaBr}\times 1000}{103g/mole\times 315g}](https://tex.z-dn.net/?f=0.565mol%2Fkg%3D%5Cfrac%7B%5Ctext%7BMass%20of%20NaBr%7D%5Ctimes%201000%7D%7B103g%2Fmole%5Ctimes%20315g%7D)
![\text{Mass of NaBr}=18.3g](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20NaBr%7D%3D18.3g)
Thus, the mass of sodium bromide added should be, 18.3 grams.
Answer:
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<span><span>m1</span>Δ<span>T1</span>+<span>m2</span>Δ<span>T2</span>=0</span>
<span><span>m1</span><span>(<span>Tf</span>l–l<span>T<span>∘1</span></span>)</span>+<span>m2</span><span>(<span>Tf</span>l–l<span>T<span>∘2</span></span>)</span>=0</span>
<span>50.0g×<span>(<span>Tf</span>l–l25.0 °C)</span>+23.0g×<span>(<span>Tf</span>l–l57.0 °C)</span>=0</span>
<span>50.0<span>Tf</span>−1250 °C+23.0<span>Tf</span> – 1311 °C=0</span>
<span>73.0<span>Tf</span>=2561 °C</span>
<span><span>Tf</span>=<span>2561 °C73.0</span>=<span>35.1 °C</span></span>
As one element .because po4 is phosphate for e.g.: NaPO4
Water is a compound that is made up of oxygen and hydrogen atoms, therefore the smallest particle of water is water molecule. The smallest particle representing water is written as H2O. H2O is the smallest unit to which water can be split while still retaining the properties of water.