Private address range network mask error

[d15k]

I've read Chapters 1-3 and got it that the private IP address range 172.16.0.0 - 172.31.255.255 has the network mask of /16. It is also being implied by review questions.

E.g., see Written lab 3.3 asks to detail the IP address 172.31.254.12 /24.
The book gives the following answer:

IP Addr Addr.class Subnet/Host bits #of subnets #of hosts
172.31.254.12 /24 B 8/8 256 254

Review question #4 in Chapter 3:

The network address of 172.16.0.0 /19 provides how many subnets and hosts?

.
The book gives the following answer:

4. F. A CIDR address of /19 is 255.255.224.0. This is a Class B address, so that is only 3 subnet
bits, but it provides 13 host bits, or 8 subnets, each with 8,190 hosts.

According to http://www.ietf.org/rfc/rfc1918.txt:

The Internet Assigned Numbers Authority (IANA) has reserved the
following three blocks of the IP address space for private internets:

10.0.0.0 - 10.255.255.255 (10/8 prefix)
172.16.0.0 - 172.31.255.255 (172.16/12 prefix)
192.168.0.0 - 192.168.255.255 (192.168/16 prefix)

We will refer to the first block as "24-bit block", the second as
"20-bit block", and to the third as "16-bit" block.

As you see the correct mask for range in subject is /12 not /16.
That means answers should be different.

Review question #4 should say: "This is a 20-bit block of private IP address address with /12 network mask (255.224.0.0), so there are 7 subnet
bits, it provides 13 host bits, or 128 subnets, each with 8,190 hosts."

Same applies for Lab 3.3.

Mr.Lammle, could you comment please?