A1.87 l aqueous solution of koh contains 155 g of koh . the solution has a density of 1.29 g/ml . calculate the molarity ( m ), molality ( m ), and mass percent concentration of the solution

[KOH] : 1.47 M

[KOH] : 1.22 m

[KOH]: 6.42 % mass percent.      

Explanation:

First of all we must determine the volume of solution. We have to work with the density

Density = mass / volume

1.29 g/ml = mass / 1870 ml

1.29 g/ml . 1870 ml = 2412.3

Now we must convert the mass to moles

155g / 56.1 g/ mol = 2.76 moles

Now we can calculate molarity

2.76 mol / 1.87 L = 1.47 M

To calculate molality we have to find out the mass of solvent

mass solute + mass solvent = mass solution

155 g + mass solvent = 2412.3 g

2412.3g - 155g = 2257.3g

We have to convert the 2257.3 g to kg

2257.3 g = 2.25 kg

molality = 2.76 moles / 2.25 kg = 1.22 m

To find out the % mass percentation, we have to calculate the mass of solute in 100 g of solution.

In 2412.3 g of solution we have 155 g of KOH

In 100 g of solution, we would have (100 . 155) / 2412.3 = 6.42 %mass percent.

answer: e.   sour taste

explanation: acids are those substances which dissociates in water to give ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. acids are sour in taste. example: hcl

bases are those substances which dissociates in water to give ions or accepts ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. bases are bitter in taste. bases are slippery in nature.

i would say b

explanation:

but i think i would need more information than that