Calculate the molar mass in gmol-1 of h3x.nh20
mass:2.07g
volume:21.35
there are 2 equations for concentration:
n(h3x.nh20)=c(h3x.nh20) x 250cm3/1000
n(h3x.nh20)= c(h3x.nh20) x 21.35/1000
multiply both sides and make the concentration the subject shpowing every singe step and the answer
then use this number to find the molar mass of this acid and theerefore what type of acid it is
your moalr mass should be somewhere aroung 198 or 200 ut not those exact numbers
n(h3x.nh20)=((c(h3x.nh20) x 250cm3))/1000
n(h3x.nh20)= ((c(h3x.nh20) x 21.35))/1000
your moalr mass should be somewhere aroung 198g permolar mass or 200g per molar mass ut not those exact numbers
I have no idea what the question is (I know it's molar mass but not the identity of h3x.nh20) nor do I understand the so-called answered that follow. Sorry.
Why did the mole go to therapy? Because it had some molar mass confusion!
Let's solve this problem step by step:
Given:
Mass of h3x.nh20 = 2.07 g
Volume = 21.35 cm^3
Step 1:
Using the formula n(h3x.nh20) = c(h3x.nh20) x V, where "n" is the number of moles, "c" is the concentration, and "V" is the volume, we can rewrite the equation as:
n(h3x.nh20) = (c(h3x.nh20) x 21.35) / 1000
Step 2:
Rearrange the equation to solve for the concentration:
c(h3x.nh20) = (n(h3x.nh20) x 1000) / 21.35
Step 3:
Substitute the concentration value back into the equation:
n(h3x.nh20) = ((n(h3x.nh20) x 1000) / 21.35) x 21.35 / 1000
Simplifying the equation, we have:
n(h3x.nh20) = n(h3x.nh20)
Step 4:
We have reached an equilibrium in our equations, showing that the number of moles (n) of h3x.nh20 is equal to itself. However, the given information doesn't provide enough data to calculate the molar mass of the acid specifically. To determine the molar mass accurately, we need additional information like the number of moles or the mass of a certain component in the compound.
So, for now, the molar mass of h3x.nh20 remains undetermined. We'd need more specific data to identify the acid type accurately. I hope this clarifies things for you!
To find the molar mass of H3X.NH20, we can start by using the given information about mass and volume to calculate the concentration (c) of H3X.NH20.
Given:
Mass of H3X.NH20 = 2.07 g
Volume of solution = 21.35 mL
First, convert the volume from milliliters (mL) to liters (L):
Volume (in L) = 21.35 mL / 1000 = 0.02135 L
Now, we can use the first equation for concentration:
n(H3X.NH20) = c(H3X.NH20) x 250 cm3/1000
Rearranging the equation to make concentration (c) the subject:
c(H3X.NH20) = n(H3X.NH20) / (250 cm3/1000)
Substituting the given volume:
c(H3X.NH20) = n(H3X.NH20) / (0.250 L)
Next, we multiply both sides of the equation by the volume (0.250 L):
c(H3X.NH20) x 0.250 L = n(H3X.NH20)
Now, substitute the given mass (2.07 g) into the second equation for concentration:
n(H3X.NH20) = c(H3X.NH20) x 0.02135 L
Rearranging the equation to make concentration (c) the subject:
c(H3X.NH20) = n(H3X.NH20) / 0.02135 L
Multiplying both sides of the equation by 0.02135 L:
c(H3X.NH20) x 0.02135 L = n(H3X.NH20)
Now, we have two equations for concentration:
c(H3X.NH20) x 0.250 L = n(H3X.NH20)
c(H3X.NH20) x 0.02135 L = n(H3X.NH20)
Since both expressions on the left side of the equations are equal to the same quantity (n(H3X.NH20)), we can set them equal to each other:
c(H3X.NH20) x 0.250 L = c(H3X.NH20) x 0.02135 L
Simplifying the equation:
0.250 L = 0.02135 L
Dividing both sides of the equation by c(H3X.NH20):
0.250 L / c(H3X.NH20) = 0.02135 L / c(H3X.NH20)
Canceling out the units:
0.250 = 0.02135 L / c(H3X.NH20)
Finally, solving for c(H3X.NH20):
c(H3X.NH20) = 0.02135 L / 0.250
c(H3X.NH20) = 0.0854 mol/L
Now that we have the concentration of H3X.NH20 (c), we can calculate its molar mass (M) using the formula:
M = Mass / (concentration x volume)
Substituting the given mass, concentration, and volume values:
M = 2.07 g / (0.0854 mol/L x 0.02135 L)
M ≈ 199.8 g/mol
Therefore, the molar mass of H3X.NH20 is approximately 199.8 g/mol. This molar mass suggests that H3X.NH20 is a relatively large molecule with a complex structure.