An aqueous 0.300 M glucose solution is prepared with a total volume of 0.150 L. The molecular weight of
1 answer:
Answer:
8.11g
Explanation:
Given parameters:
Molarity of aqueous solution = 0.3M
total volume = 0.15L
Molecular weight of glucose = 180.6g/mol
Unknown:
Mass of glucose needed in the solution = ?
Solution:
To solve this problem, we need to understand molarity.
Molarity is the number of moles of solute in a given volume of solution. In this problem, the solute here is the glucose and the solvent is water.
Molarity =
A solution is made up of solute and solvent.
now, let us solve for the number of moles of the solute which is glucose;
Number of moles of glucose = molarity x volume of solution;
= 0.3 x 0.15
= 0.045mole
Now to find the mass of glucose;
mass of glucose = number of moles x molar mass
input the parameters;
Mass of glucose = 0.045 x 180.16 = 8.11g
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Answer:
0.036 M of
Explanation:
It is an example of acid-base neutralization reaction.
KOH + ---->
+
Base Acid Salt
When two component react then the number of moles of both the component should be same, therefore the number of moles and acids and bases should be the same in the following .
Molarity=
No.of moles= Molarity × Volume of the Particular Solution
Therefore,
------------------------------(1)
where
= Molarity of Acid
= Volume of Acid
= Molarity of Base
= Volume of Base
=0.3330 M
=10.62 mL
=98.2 mL
=??(in M)
Plugging in Equation 1,
0.3330 × 10.62 = × 98.2
=
=0.036 M