Answer : The mass of NaCl required are 14.61 grams
Explanation : Given,
Molar mass of NaCl = 58.44 g/mole
Volume of solution = 500.00 mL
Molarity of NaCl = 0.5 M = 0.5 mole/L
Molarity : It is defined as the number of moles of solute present in 1 liter of solution.
Formula used :
![Molarity=\frac{\text{Mass of NaCl}\times 1000}{\text{Molar mass of NaCl}\times \text{volume of solution in ml}}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7B%5Ctext%7BMass%20of%20NaCl%7D%5Ctimes%201000%7D%7B%5Ctext%7BMolar%20mass%20of%20NaCl%7D%5Ctimes%20%5Ctext%7Bvolume%20of%20solution%20in%20ml%7D%7D)
Now put all the given values in this formula, we get:
![0.5mole/L=\frac{\text{Mass of NaCl}\times 1000}{58.44g/mole\times 500.00ml}](https://tex.z-dn.net/?f=0.5mole%2FL%3D%5Cfrac%7B%5Ctext%7BMass%20of%20NaCl%7D%5Ctimes%201000%7D%7B58.44g%2Fmole%5Ctimes%20500.00ml%7D)
![\text{Mass of NaCl}=14.61g](https://tex.z-dn.net/?f=%5Ctext%7BMass%20of%20NaCl%7D%3D14.61g)
Therefore, the mass of NaCl required are 14.61 grams
There are 0.55 moles in glucose
Answer:
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Explanation:
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1- We have to get the molar mass of calcium bromide, by getting the chemical formula ( CaBr2).
From the chemical formula, we have to get the molecular weight for each of Ca&Br on the periodic table:
Ca = 40.078 & Br = 79.904
* from the chemical formula ( CaBr2) so, the molecular weight of calcium bromide is:
40.078 + 2x(79.904) = 199.89
2- number of moles = weight / molecular weight
=20.275 / 199.89 = 0.1014 moles
3- where the volume of solution by littre = 575 mL/ 1000 = 0.575 L
So,
4-Molarity = moles of solute / L of solution
= 0.1014 / 0.575 = 0.1763 mol/L