Question

Q. If the density of methanol is 0.793 kg L-', what is its volume needed for making 2.5 L of its
0.25 M solution? pls ans fast ( NO SPAM PLS ) Otherwise I will report to SPAM answers . ​

Answer 1

Answer:

Let us calculate moles of methanol present in 2.5 L of 0.25 M solution. 

Molarity = Moles of CH3OH /Volume in L 

0.25 = Moles of CH3OH/2.5 

Moles of CH3OH = 0.25 x 2.5 = 0.625 moles

Mass of CH3OH = 0.625 x 32 = 20 g 

(Molecular mass of CH3OH = 32) 

Now 0.793 x 10³ g of CH3OH is present in 1000 mL 

20 g of CH3OH will be present in = 1000/ 0.793 x 103 x 20 = 25.2 mL 

Explanation:

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